Question

A 250-kg weight is hanging from the ceiling by a thin copper wire. In its fundamental mode, this wire at the frequency of concert A (440 Hz). You then increase the temperature of the wire by 40ºC. (a) By how much will the fundamental frequency change? Will it increase or decrease? (b) By what percentage will the speed of a wave on the wire change? (c) By what percentage will the wavelength of the fundamental standing wave change? Will it increase or decrease?

Short answer

(a) The change in fundamental frequency is -1.5Hz

(b) The percentage change of speed of wave on the wire is 0.034%

(c) The percentage change of the wavelength of fundamental standing wave is 0.068%

Step-by-step solution

 Concept of fundamental frequency.

The fundamental frequency is the minimum frequency of periodic sound waveform.

Mathematically, speed of a wave is

$\mathit{v}\mathbf{=}\sqrt{\frac{\mathbf{F}}{\mathbf{\mu }}}$

Where F is the tension force and µ is the mass per length value.

$\mathbf{v}\mathbf{=}\sqrt{\frac{\mathbf{FL}}{\mathbf{M}}}$

For fundamental frequency,

$$\begin{gathered} \mathbf{f}\mathbf{=}\frac{\mathbf{v}}{\mathbf{\lambda }}\mathbf{}\mathbf{and}\mathbf{}\mathbf{\lambda }\mathbf{=}\mathbf{2}\mathbf{\lambda } \\ \mathbf{So}\mathbf{,}\mathbf{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\sqrt{\frac{\mathbf{F}}{\mathbf{mL}}} \end{gathered}$$

(a) Determination of change in fundamental frequency.

Use differential evaluation to determine the small change in frequency corresponding to the small change in length due to thermal expansion.

$$\begin{gathered} ∆\mathrm{f}\approx \frac{\partial \mathrm{f}}{\partial \mathrm{L}}∆\mathrm{L},\mathrm{where}∆\mathrm{L}=\mathrm{L\alpha }∆\mathrm{T}\mathrm{is}\mathrm{the}\mathrm{thermal}\mathrm{expansion}\mathrm{in}\mathrm{length} \\ ∆\mathrm{f}=\frac{1}{2}\frac{1}{2}{\left(\frac{\mathrm{F}}{\mathrm{mL}}\right)}^{\frac{1}{2}}\left(\frac{\mathrm{F}}{\mathrm{m}}\right)\left(-\frac{1}{{\mathrm{L}}^2}\right)∆\mathrm{L} \\ =-\frac{1}{2}\left(\frac{1}{2}\sqrt{\frac{\mathrm{F}}{\mathrm{mL}}}\right)\frac{∆\mathrm{L}}{\mathrm{L}}=\frac{1}{2}\mathrm{f}\frac{∆\mathrm{L}}{\mathrm{L}} \\ ∆\mathrm{f}=-\frac{1}{2}\left(\mathrm{\alpha }∆\mathrm{T}\right)\mathrm{f} \\ =-\frac{1}{2}(1.7\times {10}^{-5}{\left(°\mathrm{C}\right)}^{-1})\left(40°\mathrm{C}\right)\left(440\mathrm{Hz}\right) \\ =-0.15\mathrm{Hz} \end{gathered}$$

Thus, the frequency of fundamental note decreases as length increases.

(b) Determination of percentage change of speed of wave.

Similarly, use differential evaluation to determine the change in speed,

$$\begin{gathered} ∆\mathrm{v}\approx \frac{\partial \mathrm{v}}{\partial \mathrm{L}}∆\mathrm{L} \\ \frac{∆\mathrm{v}}{\mathrm{v}}=\frac{{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{\mathrm{FL}}{\mathrm{m}}}\right)}^{-{\displaystyle \frac{1}{2}}}\left({\displaystyle \frac{\mathrm{F}}{\mathrm{m}}}\right)∆\mathrm{L}}{\sqrt{{\displaystyle \frac{\mathrm{FL}}{\mathrm{m}}}}}=\frac{∆\mathrm{L}}{2\mathrm{L}}=\frac{\mathrm{\alpha }∆\mathrm{T}}{2} \\ =\frac{1}{2}\left(1.7\times {10}^{-5}{\left(°\mathrm{C}\right)}^{-1}\right)\left(40°\mathrm{C}\right) \\ =3.4\times {10}^{-4} \\ =0.034\% \end{gathered}$$

Thus, the percentage increase in speed of wave is 0.034%

(c) Determination of percentage change in wavelength of fundamental standing wave.Taken value of wavelength is,

$$\begin{gathered} \mathrm{\lambda }=2\mathrm{L} \\ \mathrm{So},∆\mathrm{\lambda }=2∆\mathrm{L} \\ \frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}=\frac{2∆\mathrm{L}}{2\mathrm{L}}=\frac{∆\mathrm{L}}{\mathrm{L}}=\mathrm{\alpha }∆\mathrm{T} \end{gathered}$$

Substitute all the values,

$\frac{∆\mathrm{\lambda }}{\mathrm{\lambda }}=\left(1.7\times {10}^{-5}{\left(°\mathrm{C}\right)}^{-1}\right)\left(40°\mathrm{C}\right)=6.8\times {10}^{-4}=0.068\%$

Thus the percentage increase in wavelength with respect to increase in length is 0.068%.